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Posted (edited)

[math]\vec{F}=q\vec{v}\times\vec{B}[/math]

[math]\frac{d\vec{F}}{dq}=\vec{v}\times\vec{B}[/math]

[math]\int\frac{d\vec{F}}{dq} \cdot ds=\int(\frac{d\vec{s}}{dt}\times\vec{B}) \cdot ds[/math]

from here, I went about it two different ways:

1.) Here I assumed everything was at right angles and got rid of all the vectors and vector products

[math]\varepsilon=\int \frac{ds}{dt}B ds=\int \frac{ds}{dt}B \frac{ds}{dt}dt[/math]


By u substitution

[math]u=\frac{ds}{dt}, du=dt[/math]
[math]\varepsilon=\int B(u^2)du=\frac{Bv^3}{3}[/math]

where v = ds/dt


That was the first way i went about it, but i didn't feel any closer to Faraday's law.

2.) Here I left the vectors alone on the RHS; I figured since [math]\hat{v}[/math] and d[math]\hat{s}[/math] were perpendicular, the quantity ([math]\vec{v}[/math]s) would be a time derivative of the area formed

[math]\varepsilon=\int\frac{ds}{dt}B ds=\int(\vec{v}\times\vec{B}) \cdot d\vec{s}=\dot{A}B[/math]

[math]\varepsilon=\frac{BA}{dt}[/math]

don't know where the minus sign is; probably was supposed to do something with the cross product, but didn't know what.





Well I got alot further with the second "method," but is this a valid derivation? and what went wrong with the first method?

Edited by iScience92
Posted

One day I applied Lorentz's Law on a charge inside the coil of a speaker. I found that in any given instant, the force felt by a particle was always radially outward, not along the coil wire.

 

So i claim, either there is a relationship between the two laws, in which case, i'm trying to derive the less fundamental to the one more so. Or, there is another "law" of interaction going on that i don't know about, in which case, please inform me so.

Posted

 

One day I applied Lorentz's Law on a charge inside the coil of a speaker. I found that in any given instant, the force felt by a particle was always radially outward, not along the coil wire.

 

The equations governing speaker motion and their coils and driver currents are not simple electrical equations.

There is a mechanical equation to take into account as well.

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